Originally Posted by

**chiro** Hey UWM120.

So for this problem, I am going to work out the hypotenuse of the triangle where the sides are the distance of the x and y positions.

So we start off with a east distance of 230 but we need to subtract the distance based on the fact that the runner is now travelling west. So the difference squared for the x distance (i.e. find the distance and square it for pythagoras) is (230 - 7t)^2.

Now for the y-distance, we know the guy is only running west so the y-distance doesn't change. The y-distance when the runner started running is 400 so the square is 400^2.

Now the hypotenuse is just the square root of the sum of squares which is

= SQRT(400^2 - (230 - 7t)^2)